A particle in one dimension moves under the influence of a potential $V(x)= ax^6$, where $a$ is a real constant. For large $n$, what is the form of the dependence of the energy $E$ on $n$?
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For large $n$, the semiclassical approximation is valid and for bound states we may use the Bohr-Sommerfeld quantization condition: $n = \frac{1}{h}\oint p dq$, where $n$ is the principal quantum number, $h$, the Planck constant, $q$, the position and $p$, the momentum on the classical trajectory. In our case due to the conservation of energy: $\frac{p^2}{2m}+ax^6 = E \rightarrow p = \sqrt{2m(E-ax^6)}$, where $m$ is the mass and $E$, the total energy. By substitution, we obtain: $n = \frac{1}{h}\int_{-(E/a)^{1/6} }^{(E/a)^{1/6} }\sqrt{2m(E-ax^6)} dx$, where the integration is between the two turning points. By scaling the integration variable: $x =( \frac{E}{a})^{1/6}y$ We obtain: $n = \frac{\sqrt{2m}}{h}{(\frac{E}{a})}^{2/3}\int_{-1 }^{1} \sqrt{(1-y^6)} dy$. Thus: $E \propto n^{3/2}$ |
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You might not be aware, but the quantum number "n" has a classical interpretation as the action variable "J". The action variable measures the area in phase space of the classical orbit, $$J = \oint p {dx\over dt} dt$$ And the correspondence between J and n was known before quantum theory was developed. It is easy to work out the orbit shape in phase space for any value of J, and figure out what the dependence between E and J is, classically. This would solve your problem in the correspondence limit, in the limit of large n. |
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